Finding the mass of the two-dimensional object

Question

Find the mass of the two-dimensional object. A jar lid of radius 4 in with a density function of ρ(x)= 4ln(x+8).

Answer 1

Hi Gabe,

Thanks for your question!

Consider a thin ring of the jar of radius r and thickness dr. We'll call its mass dm. Since this is a two-dimensional object, mass = density * area, or symbolically,

dm = ρ(r)dA,

where ρ(r) = 4*ln(r+8) is the area density of the jar and dA is the area of the thin ring. A good approximation for the area of the thin ring is the thickness of the ring dr times the circumference of the ring 2πr. Thus, the mass of the ring becomes

dm = 2πr*ρ(r)dr.

Now we integrate over all the thin rings from r = 0 to r = 4. This will give us the total mass of the jar

m = 2π∫₀⁴r ρ(r) dr = 2π∫₀⁴r * 4ln(r+8) dr = 8π∫₀⁴r*ln(r+8) dr.

This integral can be solved using integration by parts. Let u = ln(r+8) and dv = r dr. Then, du = 1/(r+8) and v = r²/2. We have then

m = 8π∫₀⁴r*ln(r+8) dr = 8π[ (r²/2)*ln(r+8)]₀⁴ - 4π∫₀⁴r²/(r+8) dr = 64π*ln(12) - 4π∫₀⁴r²/(r+8) dr.

The integrand r²/(r+8) can be expanded as follows by polynomial long division:

r²/(r+8) = r - 8 + 64/(r+8),

so

m = 64π*ln(12) - 4π∫₀⁴r²/(r+8) dr = 64π*ln(12) - 4π∫₀⁴(r - 8) dr - 4π∫₀⁴64/(r+8)dr

= 64π*ln(12)-4π[r²/2 - 8r]₀⁴ - 256π*[ln|r+8|]₀⁴

= 64π*ln(12) - (-96π) - 256π*ln(3/2)

= 64π*ln(12) - 64π*ln(3⁴/2⁴) + 96π

= 64π*ln((2²/3)³) + 96π

= 192π*ln(4/3) + 96π

= 96π(2*ln(4/3) + 1) ≈ 475.1186.